Question

Please help!!!!
3log(x^2) + 4log(2x)=2

Answer 1

CAn you use the rules of logs to come up with a single log expression on the left side?

Answer 2

I know the first couple of steps...

Answer 3

Not sure what you're asking? Sorry.

Answer 4

Can you show the steps you already did?

Answer 5

3log(x^2) + 4log(2x)=2

log(x^2)^3 + log(2x)^4=2

log (x^6) + log (16x^4)=2

Answer 6

log (x^24) = 2?

Answer 7

The last step is not correct, but until the previous step you are correct.

Answer 8

log (x^10) = 2?

Answer 9

\(3 \log x^2 + 4 \log 2x = 2\)

\(\log (x^2)^3 + \log(2x)^4 = 2\)

\(\log x^6 + \log (2^4 x^4) = 2\)

Answer 10

Now we use the rule of logs:

\(\log a + \log b = \log (ab) \)

Answer 11

So, log (16x^24) = 2?

Answer 12

\(\log (x^6 \cdot 2^4x^4) = 2\)

\(\log (2^4 x^{10}) = 2\)

Answer 13

oh you add 6 and 4..

Answer 14

Remember, when you multiply powers with the same base, you ADD the exponents.

Answer 15

The rule is:
\(a^m \cdot a^n = a^{m + n} \)

Answer 16

So for the next step... log(2)= 16x^10?

Answer 17

Right, okay.

Answer 18

Ok. Now we use the definition of log.
Is this log base 10 or natural log?

Answer 19

log base 10

Answer 20

Definition of log:

\(\large \log_b x = y \iff b^y = x\)

Answer 21

Now let's use the last equation we have as the left side of the definition of log, and change it into the exponential version on the right side.

Answer 22

\(\log (2^4 x^{10}) = 2~~~\iff~~~10^2 = 2^4x^{10}\)

Do you understand this step?

Answer 23

Ah, yes

Answer 24

Great.
Now let's continue solving for x.

Answer 25

100=16x^10
Divide 16
6.25=x^10
6.25^(1/10)
x=1.2011

Answer 26

\(2^4x^{10} = 10^2\)

\(x^{10} = \dfrac{10^2}{2^4} \)

\(x^{10} = \left( \dfrac{10}{2^2} \right)^2\)

\(x^{10} = \left( \dfrac{5}{2} \right)^2\)

\(\left (x^{10} \right)^{\frac{1}{10}} = \left( \left( \dfrac{5}{ 2} \right)^2\ \right)^{\frac{1}{10}} \)

Answer 27

I understand now, thanks so much!

Answer 28

\(x = \left( \dfrac{5}{2} \right) ^ {\frac{1}{5}} \)

\(x = \sqrt[5] {\dfrac{5}{2} } \)

Answer 29

You're welcome.